$\text{ For }{\mathrm{CaCO}}_3\left(\mathrm{s}\right)\stackrel{\mathrm{\Delta }}{\rightleftharpoons }\mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right),\text{ the }$equilibrium constant $\left({\mathrm{K}}_{\mathrm{C}}\text{ and }{\mathrm{K}}_{\mathrm{p}}\right)$ for the thermal decomposition of calcium carbonate is

For heterogeneous chemical equilibrium,

${\mathrm{CaCO}}_3\left(\mathrm{s}\right)\stackrel{\mathrm{\Delta }}{\rightleftharpoons }\mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$

On the basis of the stoichiometric equation, we can write,

${\mathrm{K}}_{\mathrm{C}}=\frac{\left[\mathrm{CaO}\right(\mathrm{s}\left)\right]\left[{\mathrm{CO}}_2\left(\mathrm{g}\right)\right]}{\left[{\mathrm{CaCO}}_3\left(\mathrm{s}\right)\right]}$

$\text{ Since, }\left[{\mathrm{CaCO}}_3\left(\mathrm{s}\right)\right]\text{ and }\left[\mathrm{CaO}\right(\mathrm{s}\left)\right]$ are both constant, therefore modified equilibrium constant for the thermal decomposition of calcium carbonate will be

${\mathrm{K}}_{\mathrm{C}}=\left[{\mathrm{CO}}_2\left(\mathrm{g}\right)\right]\text{ or }{\mathrm{K}}_{\mathrm{p}}={\mathrm{p}}_{{\mathrm{CO}}_2}$